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Question-75248




Question Number 75248 by cesar.marval.larez@gmail.com last updated on 08/Dec/19
Commented by mathmax by abdo last updated on 14/Dec/19
(d/dx)((e^x /(1+x))) =((e^x (1+x)−e^x ×1)/((1+x)^2 )) =((xe^x )/((1+x)^2 )) ⇒  ∫  ((xe^x )/((1+x)^2 ))dx =(e^x /(1+x)) +C
$$\frac{{d}}{{dx}}\left(\frac{{e}^{{x}} }{\mathrm{1}+{x}}\right)\:=\frac{{e}^{{x}} \left(\mathrm{1}+{x}\right)−{e}^{{x}} ×\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\:=\frac{\mathrm{xe}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int\:\:\frac{{xe}^{{x}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:=\frac{{e}^{{x}} }{\mathrm{1}+{x}}\:+{C} \\ $$
Answered by mind is power last updated on 08/Dec/19
hint !xe^x =(x+1)e^x −e^x  ,    −1.e^x =−(1+x)^′ e^x
$$\mathrm{hint}\:!\mathrm{xe}^{\mathrm{x}} =\left(\mathrm{x}+\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{\mathrm{x}} \:,\:\:\:\:−\mathrm{1}.\mathrm{e}^{\mathrm{x}} =−\left(\mathrm{1}+\mathrm{x}\right)^{'} \mathrm{e}^{\mathrm{x}} \\ $$$$ \\ $$

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